# Use implicit differentiation to prove the power rule

• Oct 14th 2008, 12:54 PM
erimat89
Use implicit differentiation to prove the power rule
Let y = x^(a/b) for integers "a" and "b". Raise both sides to the "bth" power and use implicit differentiation to prove the power rule y' = (a/b)x^(a/b-1).
• Oct 14th 2008, 01:04 PM
vincisonfire
y^b=x^a
b*y^(b-1)*y'=a*x^(a-1)
y'={a*x^(a-1)}/{b*y^(b-1)}
y'=(a/b)*x^(a-1)/x^(a*(b-1)/b)
y' = (a/b)x^(a/b-1)
• Oct 14th 2008, 01:32 PM
erimat89
Confused
What exactly happened to the b*y in the 3rd step to the 4th step? Why did the y variable disappear?
• Oct 14th 2008, 01:34 PM
Soroban
Hello, erimat89!

I'll do it in LaTex . . .

Quote:

Let $\displaystyle y \:=\: x^{\frac{a}{b}}$ for integers $\displaystyle a\text{ and }b$
Raise both sides to the power $\displaystyle b$ and use implicit differentiation
to prove the power rule: .$\displaystyle y' \:=\:\frac{a}{b}\!\cdot\!x^{\frac{a}{b}-1}$

We have: .$\displaystyle y \;=\;x^{\frac{a}{b}}$ .[1]

Raise to the $\displaystyle b^{th}$ power: .$\displaystyle y^b \:=\:x^a$ .[2]

Differentiate implicitly: .$\displaystyle by^{b-1}y' \;=\;ax^{a-1} \quad\Rightarrow\quad y' \;=\;\frac{a}{b}\cdot\frac{x^{a-1}}{y^{b-1}}$ .[3]

Divide [2] by $\displaystyle y\!:\;\;y^{b-1} \:=\:\frac{x^a}{y}$

Substitute [1]: .$\displaystyle y^{b-1} \:=\: \frac{x^a}{x^{\frac{a}{b}}} \:=\:x^{a-\frac{a}{b}}$

Substitute into [3]: .$\displaystyle y' \;=\;\frac{a}{b}\cdot\frac{x^{a-1}}{x^{a-\frac{a}{b}}} \quad\Rightarrow\quad\boxed{ y' \;=\;\frac{a}{b}\!\cdot\!x^{\frac{a}{b}-1}}$

• Oct 14th 2008, 01:34 PM
vincisonfire